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Need to add Equivalent conditions to a map being a quotient map

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Definition

There are a few definitions of the quotient topology however they do not conflict. This page might change shape while things are put in place

Quotient topology via an equivalence-relation definition

Given a topological space, [ilmath](X,\mathcal{J})[/ilmath] and an equivalence relation on [ilmath]X[/ilmath], [ilmath]\sim[/ilmath]^{[Note 1]}, the quotient topology on [ilmath]\frac{X}{\sim} [/ilmath], [ilmath]\mathcal{K} [/ilmath] is defined as:

The set [ilmath]\mathcal{K}\subseteq\mathcal{P}(\frac{X}{\sim})[/ilmath] such that:
- [ilmath]\forall U\in\mathcal{P}(\frac{X}{\sim})[U\in\mathcal{K}\iff \pi^{-1}(U)\in\mathcal{J}][/ilmath] or equivalently
[ilmath]\mathcal{K}=\{U\in\mathcal{P}(\frac{X}{\sim})\ \vert\ \pi^{-1}(U)\in\mathcal{J}\}[/ilmath]

In words:

The topology on [ilmath]\frac{X}{\sim} [/ilmath] consists of all those sets whose pre-image (under [ilmath]\pi[/ilmath]) are open in [ilmath]X[/ilmath]

Claim 1: [ilmath]\mathcal{K} [/ilmath] is indeed a topology on [ilmath]\frac{X}{\sim} [/ilmath]

Quotient topology via a mapping to a set definition

Let [ilmath](X,\mathcal{J})[/ilmath] be a topological space and let [ilmath]h:X\rightarrow Y[/ilmath] be a surjective map onto a set [ilmath]Y[/ilmath], then the quotient topology, [ilmath]\mathcal{K}\subseteq\mathcal{P}(Y)[/ilmath] is a topology we define on [ilmath]Y[/ilmath] as follows:

[ilmath]\forall U\in\mathcal{P}(Y)[Y\in\mathcal{K}\iff h^{-1}(U)\in\mathcal{J}][/ilmath] or equivalently:
[ilmath]\mathcal{K}=\{U\in\mathcal{P}(Y)\ \vert\ h^{-1}(U)\in\mathcal{J}\}[/ilmath]

The quotient topology on [ilmath]Y[/ilmath] consists of all those subsets of [ilmath]Y[/ilmath] whose pre-image (under [ilmath]h[/ilmath]) is open in [ilmath]X[/ilmath]

Claim 2: these definitions are equivalent

Immediate theorems

The next two theorems demonstrate the purpose, the job if you will, of the quotient topology. The second (passing to the quotient) is the most important.

Universal property of the quotient topology

In this commutative diagram [ilmath]f[/ilmath] is continuous [ilmath]\iff[/ilmath] [ilmath]f\circ q[/ilmath] is continuous
[ilmath]\xymatrix{ X \ar[d]_{q} \ar[dr]^{f\circ q} & \\ Y \ar[r]_f & Z }[/ilmath]

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]q:X\rightarrow Y[/ilmath] be a quotient map. Then^[1]:

For any topological space, [ilmath](Z,\mathcal{ H })[/ilmath] a map, [ilmath]f:Y\rightarrow Z[/ilmath] is continuous if and only if the composite map, [ilmath]f\circ q[/ilmath], is continuous

Passing to the quotient

[ilmath]f[/ilmath] descends to the quotient
[ilmath]\xymatrix{ X \ar[d]_\pi \ar[dr]^f & \\ \frac{X}{\sim} \ar@{.>}[r]^{\overline{f} }& Y}[/ilmath]

Suppose that [ilmath](X,\mathcal{ J })[/ilmath] is a topological space and [ilmath]\sim[/ilmath] is an equivalence relation, let [ilmath](\frac{X}{\sim},\mathcal{ Q })[/ilmath] be the resulting quotient topology and [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] the resulting quotient map, then:

Let [ilmath](Y,\mathcal{ K })[/ilmath] be any topological space and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map that is constant on the fibres of [ilmath]\pi[/ilmath]^{[Note 2]} then:
there exists a unique continuous map, [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] such that [ilmath]f=\overline{f}\circ\pi[/ilmath]

We may then say [ilmath]f[/ilmath] descends to the quotient or passes to the quotient

Note: this is an instance of passing-to-the-quotient for functions

Proof of claims

Notes

↑ Recall that for an equivalence relation there is a natural map that sends each [ilmath]x\in X[/ilmath] to [ilmath][x][/ilmath] (the equivalence class containing [ilmath]x[/ilmath]) which we denote here as [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath]. Recall also that [ilmath]\frac{X}{\sim} [/ilmath] denotes the set of all equivalence classes of [ilmath]\sim[/ilmath].
↑
That means that:
- [ilmath]\forall x,y\in X[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath] - as mentioned in passing-to-the-quotient for functions, or
- [ilmath]\forall x,y\in X[f(x)\ne f(y)\implies \pi(x)\ne\pi(y)][/ilmath], also mentioned
- See :- Equivalent conditions to being constant on the fibres of a map for details

References

↑ Introduction to Topological Manifolds - John M. Lee

OLD PAGE

Note: Motivation for quotient topology may be useful

Definition of the Quotient topology

[math] \begin{xy} \xymatrix{ X \ar[r]^p \ar[dr]_f & Q \ar@{.>}[d]^{\tilde{f}}\\ & Y } \end{xy} [/math]

(OLD)Definition of Quotient topology

If [math](X,\mathcal{J})[/math] is a topological space, [math]A[/math] is a set, and [math]p:(X,\mathcal{J})\rightarrow A[/math] is a surjective map then there exists exactly one topology [math]\mathcal{J}_Q[/math] relative to which [math]p[/math] is a quotient map. This is the quotient topology induced by [math]p[/math]

The quotient topology is actually a topology

TODO: Easy enough

Quotient map

Let [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath] be topological spaces and let [ilmath]p:X\rightarrow Y[/ilmath] be a surjective map.

[ilmath]p[/ilmath] is a quotient map^[1] if we have [math]U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}[/math]

That is to say [math]\mathcal{K}=\{V\in\mathcal{P}(Y)|p^{-1}(V)\in\mathcal{J}\}[/math]

Also known as:

Identification map

Stronger than continuity

If we had [ilmath]\mathcal{K}=\{\emptyset,Y\}[/ilmath] then [ilmath]p[/ilmath] is automatically continuous (as it is surjective), the point is that [ilmath]\mathcal{K} [/ilmath] is the largest topology we can define on [ilmath]Y[/ilmath] such that [ilmath]p[/ilmath] is continuous

Theorems

Theorem: The quotient topology, [ilmath]\mathcal{Q} [/ilmath] is the largest topology such that the quotient map, [ilmath]p[/ilmath], is continuous. That is to say any other topology such on [ilmath]Y[/ilmath] such that [ilmath]p[/ilmath] is continuous is contained in the quotient topology

For a map [ilmath]p:X\rightarrow Y[/ilmath] where [ilmath](X,\mathcal{J})[/ilmath] is a Topological space we will show that the topology on [ilmath]Y[/ilmath] given by:

[math]\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}[/math]

is the largest topology on [ilmath]Y[/ilmath] we can have such that [ilmath]p[/ilmath] is continuous

Proof method: suppose there's a larger topology, reach a contradiction.

Suppose that [ilmath]\mathcal{K} [/ilmath] is any topology on [ilmath]Y[/ilmath] and that [ilmath]p:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})[/ilmath] is continuous.

Suppose that [ilmath]\mathcal{K}\ne\mathcal{Q} [/ilmath]

Let [ilmath]V\in\mathcal{K} [/ilmath] such that [ilmath]V\notin \mathcal{Q} [/ilmath]

By continuity of [ilmath]p[/ilmath], [ilmath]p^{-1}(V)\in\mathcal{J} [/ilmath]

This contradicts that [ilmath]V\notin\mathcal{Q} [/ilmath] as [ilmath]\mathcal{Q} [/ilmath] contains all subsets of [ilmath]Y[/ilmath] whose inverse image (preimage) is open in [ilmath]X[/ilmath]

Thus any topology on [ilmath]Y[/ilmath] where [ilmath]p[/ilmath] is continuous is contained in the quotient topology

This theorem hints at the Characteristic property of the quotient topology

Quotient space

Given a Topological space [ilmath](X,\mathcal{J})[/ilmath] and an Equivalence relation [ilmath]\sim[/ilmath], then the map: [math]q:(X,\mathcal{J})\rightarrow(\tfrac{X}{\sim},\mathcal{Q})[/math] with [math]q:p\mapsto[p][/math] (which is a quotient map) is continuous (as above)

The topological space [ilmath](\tfrac{X}{\sim},\mathcal{Q})[/ilmath] is the quotient space^[2] where [ilmath]\mathcal{Q} [/ilmath] is the topology induced by the quotient

Also known as:

Identification space

References

↑ Topology - Second Edition - James R Munkres
↑ Introduction to topological manifolds - John M Lee - Second edition

[1] Recall that for an equivalence relation there is a natural map that sends each [ilmath]x\in X[/ilmath] to [ilmath][x][/ilmath] (the equivalence class containing [ilmath]x[/ilmath]) which we denote here as [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath]. Recall also that [ilmath]\frac{X}{\sim} [/ilmath] denotes the set of all equivalence classes of [ilmath]\sim[/ilmath].

[3] That means that:
[ilmath]\forall x,y\in X[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath] - as mentioned in passing-to-the-quotient for functions, or

[ilmath]\forall x,y\in X[f(x)\ne f(y)\implies \pi(x)\ne\pi(y)][/ilmath], also mentioned

See :- Equivalent conditions to being constant on the fibres of a map for details

[3] [ilmath]\forall x,y\in X[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath] - as mentioned in passing-to-the-quotient for functions, or

[4] [ilmath]\forall x,y\in X[f(x)\ne f(y)\implies \pi(x)\ne\pi(y)][/ilmath], also mentioned

[5] See :- Equivalent conditions to being constant on the fibres of a map for details

[ITTMJML-2] Introduction to Topological Manifolds - John M. Lee

[4] Topology - Second Edition - James R Munkres

[5] Introduction to topological manifolds - John M Lee - Second edition

[Note 1]

[1]

[Note 2]

[1]

[2]

@@ Line 1: / Line 1: @@
+* Need to add [[Equivalent conditions to a map being a quotient map]]
+{{Refactor notice|grade=A|msg=This page is an embarrassment}}
 ==Definition==
+There are a few definitions of the quotient topology however they do not conflict. This page might change shape while things are put in place
+===[[Quotient topology/Equivalence relation definition|Quotient topology via an equivalence-relation definition]]===
+{{:Quotient topology/Equivalence relation definition}}
+: '''Claim 1: ''' {{M|\mathcal{K} }} is indeed a topology on {{M|\frac{X}{\sim} }}
+===[[Quotient topology/Mapping to a set definition|Quotient topology via a mapping to a set definition]]===
+{{:Quotient topology/Mapping to a set definition}}
+: '''Claim 2: ''' these definitions are equivalent
+==Immediate theorems==
+The next two theorems demonstrate the purpose, the job if you will, of the quotient topology. The second (''[[passing to the quotient (topology)|passing to the quotient]]'') is the most important.
+{{Begin Inline Theorem}}
+===[[Universal property of the quotient topology]]===
+{{Begin Inline Proof}}
+{{:Universal property of the quotient topology/Statement}}
+{{End Proof}}{{End Theorem}}
+{{Begin Inline Theorem}}
+===[[Passing to the quotient (topology)|Passing to the quotient]]===
+{{Begin Inline Proof}}
+{{:Passing to the quotient (topology)/Statement}}
+{{End Proof}}{{End Theorem}}
+==Proof of claims==
+==Notes==
+<references group="Note"/>
+==References==
+<references/>
+{{Topology navbox|plain}}
+{{Definition|Topology}}
+=OLD PAGE=
+'''Note:''' [[Motivation for quotient topology]] may be useful
+==Definition of the Quotient topology==
+<math>
+\begin{xy}
+\xymatrix{
+X \ar[r]^p \ar[dr]_f & Q \ar@{.>}[d]^{\tilde{f}}\\
+ & Y
+}
+\end{xy}
+</math>
+==(OLD)Definition of Quotient topology==
 If <math>(X,\mathcal{J})</math> is a [[Topological space|topological space]], <math>A</math> is a set, and <math>p:(X,\mathcal{J})\rightarrow A</math> is a [[Surjection|surjective map]] then there exists '''exactly one''' topology <math>\mathcal{J}_Q</math> relative to which <math>p</math> is a quotient map. This is the '''quotient topology''' induced by <math>p</math>
-{{Todo|Munkres page 138}}
+{{Begin Theorem}}
+The quotient topology is actually a topology
+{{Begin Proof}}
+{{Todo|Easy enough}}
+{{End Proof}}
+{{End Theorem}}
+==Quotient map==
+Let {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}} be [[Topological space|topological spaces]] and let {{M|p:X\rightarrow Y}} be a [[Surjection|surjective]] map.
+{{M|p}} is a quotient map<ref>Topology - Second Edition - James R Munkres</ref> if we have <math>U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}</math>
+That is to say <math>\mathcal{K}=\{V\in\mathcal{P}(Y)|p^{-1}(V)\in\mathcal{J}\}</math>
+Also known as:
+* Identification map
+===Stronger than continuity===
+If we had {{M|1=\mathcal{K}=\{\emptyset,Y\} }} then {{M|p}} is automatically continuous (as it is surjective), the point is that {{M|\mathcal{K} }} is the [[Topology#Finer.2C_Larger.2C_Stronger|largest topology]] we can define on {{M|Y}} such that {{M|p}} is continuous
+===Theorems===
+{{Begin Theorem}}
+Theorem: The quotient topology, {{M|\mathcal{Q} }} is the largest topology such that the quotient map, {{M|p}}, is continuous. That is to say any other topology such on {{M|Y}} such that {{M|p}} is continuous is contained in the quotient topology
+{{Begin Proof}}
+For a map {{M|p:X\rightarrow Y}} where {{M|(X,\mathcal{J})}} is a [[Topological space]] we will show that the topology on {{M|Y}} given by:
+* <math>\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}</math>
+is the largest topology on {{M|Y}} we can have ''such that'' {{M|p}} is [[Continuous map|continuous]]
+'''Proof method:''' suppose there's a larger topology, reach a contradiction.
+Suppose that {{M|\mathcal{K} }} is any topology on {{M|Y}} and that {{M|p:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})}} is continuous.
+Suppose that {{M|\mathcal{K}\ne\mathcal{Q} }}
+Let {{M|V\in\mathcal{K} }} such that {{M|V\notin \mathcal{Q} }}
+By continuity of {{M|p}}, {{M|p^{-1}(V)\in\mathcal{J} }}
+This contradicts that {{M|V\notin\mathcal{Q} }} as {{M|\mathcal{Q} }} contains all subsets of {{M|Y}} whose inverse image (preimage) is open in {{M|X}}
+Thus any topology on {{M|Y}} where {{M|p}} is continuous is contained in the quotient topology
+{{End Proof}}
+{{End Theorem}}
+This theorem hints at the [[Characteristic property of the quotient topology]]
+==Quotient space==
+Given a [[Topological space]] {{M|(X,\mathcal{J})}} and an [[Equivalence relation]] {{M|\sim}}, then the map: <math>q:(X,\mathcal{J})\rightarrow(\tfrac{X}{\sim},\mathcal{Q})</math> with <math>q:p\mapsto[p]</math> (which is a quotient map) is continuous (as above)
+The topological space {{M|(\tfrac{X}{\sim},\mathcal{Q})}} is the ''quotient space''<ref>Introduction to topological manifolds - John M Lee - Second edition</ref> where {{M|\mathcal{Q} }} is the topology induced by the quotient
+Also known as:
+* Identification space
+==See also==
+* [[Open and closed maps]]
+==References==
+<references/>
 {{Definition|Topology}}

Difference between revisions of "Quotient topology"

Latest revision as of 13:34, 16 October 2016

Contents

Definition

Quotient topology via an equivalence-relation definition

Quotient topology via a mapping to a set definition

Immediate theorems

Universal property of the quotient topology

Passing to the quotient

Proof of claims

Notes

References

OLD PAGE

Definition of the Quotient topology

(OLD)Definition of Quotient topology

Quotient map

Stronger than continuity

Theorems

Quotient space

See also

References

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